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For which $n\geq0$ is $n!>2^n$ valid?

My approach by induction:

Consider the base case $n=0$:

$0!\ngtr2^0$

Let us consider $n+1$ case:

$$(n+1)!>2^{n+1}$$ $$(n+1)n!>2^{n}\cdot 2$$ $$\frac{n+1}{2^n}>\frac{2}{n!}$$

Let us evaluate the limits of the LHS and RHS:

$$\lim_{n\longrightarrow\infty}\frac{n+1}{2^n}=0$$ $$\lim_{n\longrightarrow\infty}\frac{2}{n!}=0$$

since the RHS term approximates the limit faster, then we can consider the case of the denominators by induction:

$$2^n>n!$$ and evaluate the first given values of $n=1,2,3,\cdots,5$

$$2^1>1!, \ \text{which holds}$$ $$2^2>2!, \ \text{which does not hold}$$ $$2^3>3!, \ \text{which holds}$$ $$2^4>4!, \ \text{which does not hold}$$ $$2^5>5!, \ \text{which does not hold}$$ $$2^{5+n}>(5+n)!, \ \text{which does not hold}$$ So we have found that $n!>2^n$ holds for all $n\ne \{0,2,4,5,(5+k)\}$, where $k=1,2,\cdots $

But something is not right here, apparently the limit and the final statement. But which other ways can it be solved?

Thanks

Bill Dubuque
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Superunknown
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    By inspection, $~n! \leq 2^n,~$ for $~n \in {0,1,2,3},~$ and $~4! > 2^4.~$ Further, $$[~n > 1 ~~\text{and} ~~n! > 2^n~] \implies (n+1)! > (n+1)2^n \geq 2^{n+1}.$$ – user2661923 Jul 06 '23 at 09:24
  • @user2661923 why do you use $\leq$ sign, when it the OP considers $<$ sign? – Superunknown Jul 07 '23 at 08:34
  • @user2661923

    $$2^1>1!, \ \text{which holds}$$ $$2^2>2!, \ \text{which does not hold}$$ $$2^3>3!, \ \text{which holds}$$ $$2^4>4!, \ \text{which does not hold}$$ $$2^5>5!, \ \text{which does not hold}$$ $$2^{5+n}>(5+n)!, \ \text{which does not hold}$$

     – Superunknown Jul 07 '23 at 08:48
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    Because $~0!~$ is not less than $~2^0.$ – user2661923 Jul 07 '23 at 14:21
  • @user2661923 Does this mean that the actual question is inaccurate? – Superunknown Jul 07 '23 at 16:53
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    No. The question asks for which $~n \in \Bbb{Z^+}~$ is $~n! > 2^n.~$ The answer is $~n \in \Bbb{Z_{\geq 4}}.$ – user2661923 Jul 07 '23 at 23:32

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